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3 years ago

There are 3/4 of a pizza left over from last night how many 1/8 portions are there

Mathematics

1 answer:

KiRa [710]3 years ago

8 0

3/4 of the pizza is left

3/4 = 6/8

There are 6 "1/8" portions left

hope this helps

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Suppose that rectangle ABCD is dilated to A'B'C'D' by a scale factor of 0.5 with a center of dilation at (4, 6). What is the dis

Rasek [7]

To solve this problem you must apply the proccedure shown below:

1. You must multiply the coordinates of $BC$ by the scale factor given in the problem above. Therefore, you have that the coordinates of $B'C'$ are:

$B'(1,5)\\ C'(3,5)$

2. The midpoint of $B'C'$ has the folllowing coordinates:

$m(\frac{x1+x2}{2} , \frac{y1+y2}{2} )\\ m(\frac{1+3}{2}, \frac{5+5}{2})\\ m(2,5)$

3. The center of dilation is a fixed point, with coordinates $(4,6)$ . Therefore, the distance from the center of dilation to the midpoint of $B'C'$ is:

$distance=x2-x1\\distance=4 units-2units\\distance=2 units$

The answer is the option C: $2 units$

5 0

3 years ago

Read 2 more answers

9.) express the area of this triangle as a polynomial. (Remember A=1/2B H

schepotkina [342]

Plug in the values for b = $4ab^2c^7$ and h = $3a^2b^5$

A = $\frac{1}{2} (4ab^2c^7)(3a^2b^5)$
Simplify....
A = $6a^3b^7c^7$

5 0

3 years ago

Read 2 more answers

Please help meeee i beg u

mixer [17]

Answer:

its c because b is a right angle triangle

thats how i solved it if there is a mistake tell me

4 0

4 years ago

Suppose that a function f(x) is defined for all real values of x except at xequals=c. can anything be said about the existence

Margaret [11]

we are given that

f(x) is defined for all values of x except at x=c

Limit may or may not exist

case-1:

If there is hole at x=c , then limit exist

case-2:

If there is vertical asymptote at x=c , then limit does not exist

Examples:

case-1:

$\lim_{x \to c} \frac{x^2-cx}{(x-c)}$

We can simplify it

$\lim_{x \to c} \frac{x(x-c)}{(x-c)}$

$=\lim_{x \to c} x$

$=c$

so, we can see that limit exist and it's value defined

case-2:

$\lim_{x \to c} \frac{1}{(x-c)}$

Left limit is

$\lim_{x \to c-} \frac{1}{(x-c)}$

$=-\infty$

Right Limit is

$\lim_{x \to c+} \frac{1}{(x-c)}$

$=+\infty$

so, we can see that left limit is not equal to right limit

so, limit does not exist

4 0

3 years ago

Calculate the area of the following figure correct to 2 decimal places if can show or explain work

grin007 [14]

The way we approach a question like this, is by looking at what portion of the circle we actually have to calculate the area of. We know that a circle is a shape that goes 360 degrees around a point. Therefore we know that, for example, a semicircle, or half a circle, would cover half the amount of degrees. This is important, because as well as covering half the number of degrees, it also covers half the area. For the figure in your picture, we are told that the portion missing from the circle is 90 degrees (shown by the right angle symbol in the center).
this means that 90 degrees worth of the circle is missing, so if we do 360÷90, we find that the answer equals $\frac{3}{4}$ . This means that we are dealing with $\frac{3}{4}$ of a circle. So, we simply find the area of the circle normally, and then multiply by <span> $\frac{3}{4}$ :
The formula for area of a circle is </span> $\pi r^{2}$ , so in this case, it can be said to be $\pi (10^{2} )$ , which is equal to 314.15
Now we simply take that area and multiply it by the fraction we established earlier:
314.15 x $\frac{3}{4}$ = 235.61
Therefore the area of the figure is 235.61 $cm^{2}$
I hope this helped and please remember to try and not only understand the answer, but the maths that worked it out too :))

3 0

4 years ago